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W A gas is compressed at a constant pressure of \(0.800 \mathrm{~atm}\) from \(9.00 \mathrm{~L}\) to \(2.00 \mathrm{~L}\). In the process, \(400 \mathrm{~J}\) of energy leaves the gas by heat. (a) What is the work done on the gas? (b) What is the change in its internal energy?

Short Answer

Expert verified
The work done on the gas and the change in its internal energy can be found by substituting the given values into the equations mentioned in the steps. The work done on the gas and the change in its internal energy are negative.

Step by step solution

01

Calculate Work Done

The formula to calculate work done at constant pressure is \(W = P \Delta V\) where \(\Delta V\) is the change in volume. Since the work is done on the gas, work done is negative. So, \(\Delta V = V_{initial} - V_{final} = 9.00 \mathrm{~L} - 2.00 \mathrm{~L} = -7.00 \mathrm{~L} = -7.00 \times 10^{-3} \mathrm{~m^3} \, (since 1 \mathrm{~L} = 10^{-3} \mathrm{~m^3}) \). Pressure \( P = 0.800 \, \mathrm{atm} = 0.800 \times 101325 \, \mathrm{N/m^2} \, (since 1 \, \mathrm{atm} = 101325 \, \mathrm{N/m^2}) \). Then calculate \( W = P \Delta V \) .
02

Calculate Change in Internal Energy

The first law of thermodynamics states that for an isolated system, the change in internal energy is equal to the heat added to the system minus the work done by the system. That is, \(\Delta U = Q - W\), where \(Q\) is the heat added to the system and \(W\) is the work done by the system. If the heat leaves the system, \(Q\) is negative. Here, \(400 \, \mathrm{J}\) of energy leaves the gas by heat, so \(Q = -400\, \mathrm{J}\) and the work done on the gas (found in step 1) should be regarded as negative. Now, substitute the values of \(Q\) and \(W\) into the equation \(\Delta U = Q - W\) to find \(\Delta U\).
03

Short Answer

Evaluate the equations from steps 1 and 2 to get the work done on the gas and the change in its internal energy. The answer should be in Joules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done in Thermodynamics
In thermodynamics, 'work done' is a measure of energy transfer when an object is moved by an applied force. Specifically in the context of gases, work is done when the gas either expands or is compressed by external pressure. This is often represented with the formula
\( W = P \times \text{{\textDelta}}V \)
, where \( W \) is the work, \( P \) is the constant external pressure, and \( \text{{\textDelta}}V \) is the change in volume of the gas.

For example, when a gas is compressed at constant pressure, the work done on the gas is considered to be negative, because the gas has work done on it (it doesn't expand to do work on the surroundings). Therefore, if a gas is compressed from a larger volume to a smaller one, the formula accounts for this by a negative change in volume, \( \text{{\textDelta}}V = V_{\text{{initial}}} - V_{\text{{final}}} \). This reflects the idea that the internal energy of the gas increases, as work is being done on it rather than by it.
First Law of Thermodynamics
The first law of thermodynamics, which is a version of the law of conservation of energy, states that the energy within a closed system remains constant over time. According to this principle, energy within the system can be transformed in form or transported in or out of the system, but it can never be created or destroyed. Mathematically, it's expressed as
\( \text{{\textDelta}}U = Q - W \)
, where \( \text{{\textDelta}}U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. If work is done on the system, it's input as negative, and if heat is leaving the system, this is also considered negative. This law is at the heart of all thermal processes and understanding it is crucial for solving problems related to the conservation of energy within thermodynamic systems.

Applying this to practical scenarios, if we consider a gas within a piston being compressed, it may lose some energy as heat but gain internal energy from the work done on it, and according to the first law, we can calculate the change in the gas's internal energy.
Internal Energy Change
The concept of 'internal energy' relates to the total energy contained within a system, attributed to the kinetic and potential energy of its molecules. The change in internal energy (\( \text{{\textDelta}}U \)) of a system can be calculated from the first law of thermodynamics as mentioned, considering the heat added to or lost from the system and the work done by or on the system.

When work is done on the gas, as is the case with compression against a constant pressure, the internal energy of the gas increases. On the other hand, if the gas expands and does work on its surroundings, it loses internal energy. Any heat transfer also affects internal energy: adding heat increases it, while losing heat decreases it. Grasping this change is essential for understanding how thermal systems operate, as it affects the system's temperature and can carry implications for phase changes or chemical reactions occurring within the system.
Constant Pressure Process
A 'constant pressure process', also known as an isobaric process, occurs when a thermodynamic process happens at unvarying pressure. It’s particularly relevant in various practical applications such as heating or cooling within closed systems. In these cases, the work done by or on the system can be computed straightforwardly by the earlier mentioned formula \( W = P \times \text{{\textDelta}}V \), where \( P \) is maintained constant throughout the entire process.

An example in everyday life is a piston engine, where the expansion or compression of gases occurs at or close to constant pressure. Understanding constant pressure processes is critical for students studying thermodynamics because it helps elucidate how the essential properties of the system such as volume, temperature, and internal energy will change in response to work being done.

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Most popular questions from this chapter

A certain nuclear power plant has an electrical power output of \(435 \mathrm{MW}\). The rate at which energy must be supplicd to the plant is \(1420 \mathrm{MW}\). (a) What is the thermal efficiency of the power plant? (b) At what rate is thermal energy expelled by the plant?

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A heat engine operates between a reservoir at \(25^{\circ} \mathrm{C}\) and one at \(375^{\circ} \mathrm{C}\). What is the maximum efficiency possible for this engine?

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An ideal monatomic gas expands isothermally from \(0.500 \mathrm{~m}^{3}\) to \(1.25 \mathrm{~m}^{3}\) at a constant temperature of \(675 \mathrm{~K}\). If the initial pressure is \(1.00 \times 10^{5} \mathrm{~Pa}\), find (a) the work done on the gas, (b) the thermal energy transfer \(Q\), and (c) the change in the internal energy.

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